Question

A rock is thrown upward from the edge of a bridge and onto a road that is 10 feet below the bridge function h(x)=-x^2+3x+10 gives the height, h, in feet, the rock travels in x seconds from the time it was thrown. When will the rock hit the road?​

Answer 1

Answer:

Rock will hit the road after 5 seconds.

Step-by-step explanation:

The given function is

$h(x)=-x^2+3x+10$

where, h is the height of the rock after x seconds.

The height of rock is 0 if it hits the road.

Substitute h(x)=0 in the given function.

$0=-x^2+3x+10$

Splitting the middle term we get

$0=-x^2+(5x-2x)+10$

$0=(-x^2+5x)+(-2x+10)$

On further simplification we get

$0=-x(x-5)-2(x-5)$

$0=(x-5)(-x-2)$

$0=-(x-5)(x+2)$

Using zero product property.

$x-5=0\Rightarrow x=5$

$x+2=0\Rightarrow x=-2$

Time can not be negative. So, the reasonable value of x is 5.

Therefore, Rock will hit the road after 5 seconds.

Answer 2

Answer:

after 5 seconds

Step-by-step explanation:

Given

h(x) = - x² + 3x + 10

When the rock hits the road the height h = 0

Equate h(x) to zero and solve for x, that is

- x² + 3x + 10 = 0 ( multiply through by - 1 )

x² - 3x - 10 = 0 ← in standard form

(x - 5)(x + 2) = 0 ← in factored form

Equate each factor to zero and solve for x

x - 5 = 0 ⇒ x = 5

x + 2 = 0 ⇒ x = - 2

However, x > 0 ⇒ x = 5

Thus the rock hits the road after 5 seconds.